All the pandigitals are multiples of 9, so the question is how many are multiples of 11.

If the even digits are as indicated below for those cases where the sum of the even and that of the odd digits differ by a multiple of 11, the sum of the even digits and that of the odd digits are as shown on the same line:

In 11 of the cases there is a zero in the odd positions, including the first position. In each of those cases, there are 5! - 4! = 96 valid permutations of the odd digits and 5! = 120 of the even digits for 96*120=11520 valid permutations of all the digits.

In the other 11 cases, there are 5! valid permutations of each. Then, 120^2 = 14400 is the number of valid permutations of all the digits.